1408 CHAPTER 42. MAXIMAL MONOTONE OPERATORS, HILBERT SPACE
Now the first term in 42.9.65 is bounded so consider the second. The integral in this termis of the form∫ t
0e
sλ φ (s,z(s))ds+
∫ t
0e
sλ (φ (t,z(s))−φ (s,z(s)))ds. (42.9.66)
Since [z] ∈ D(Φ) , φ (s,z(s))< ∞ for all s and also the first integral in 42.9.66 is finite. By42.9.59, the second term in 42.9.66 is dominated by
Cλ
∫ t
0K(
1+φ (s,z(s))+ |z(s)|2)|t− s|ds < ∞.
This shows φ (t,Jλ [z] (t))< ∞ for all t and so Φ([Jλ [z]]) is given by the top line of 42.9.61.Therefore, by convexity of φ (t, ·) and Jensen’s inequality,
Φ([Jλ [z]]) =∫ T
0φ
(t,e
−tλ x0 +
1λ
e−tλ
∫ t
0e
sλ z(s)ds
)dt
≤∫ T
0
(e−tλ φ (t,x0)+
1λ
e−tλ
∫ t
0e
sλ φ (t,z(s))ds
)dt
=∫ T
0e−tλ φ (t,x0)dt +
∫ T
0
1λ
e−tλ
∫ t
0e
sλ φ (s,z(s))dsdt
+∫ T
0
1λ
e−tλ
∫ t
0e
sλ (φ (t,z(s))−φ (s,z(s)))dsdt. (42.9.67)
By 42.9.59, the last term is dominated by∫ T
0
∫ t
0
1λ
e−(t−s)
λ K(
1+φ (s,z(s))+ |z(s)|2)|t− s|dsdt =
∫ T
0
∫ T
se−(t−s)
λ
t− sλ
dtK(
1+φ (s,z(s))+ |z(s)|2)
ds
≤Cλ +Cλ
(Φ([z])+ |[z]|2
). (42.9.68)
for some constant, C. From 42.9.59, φ (t,x0) is bounded and so the first term in 42.9.67is dominated by an expression of the form Cλ . Now consider the middle term of 42.9.67.Since φ is nonnegative,∫ T
0
1λ
e−tλ
∫ t
0e
sλ φ (s,z(s))dsdt =
∫ T
0
∫ T
s
1λ
e−(t−s)
λ dtφ (s,z(s))ds
≤∫ T
0
∫∞
0e−uduφ (s,z(s))ds = Φ([z]) . (42.9.69)
It followsΦ([Jλ [z]])≤Φ([z])+Cλ +Cλ
(Φ([z])+ |[z]|2
).